Removing the interior of a neighbourhood of the core curve of solid tori
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I have given a lens space L(p,q). What compact three-manifold with boundary do we get if we remove the interior of a neighbourhood of the core curve of one of the two solid tori that make up the lens space?
I figured out that I have to remove a solid torus from L(p,q) but I can not visulaize or proof what do I get if do that.
general-topology knot-theory
asked Dec 12 '18 at 13:50
ValVal
1
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add a comment |
$begingroup$
I have given a lens space L(p,q). What compact three-manifold with boundary do we get if we remove the interior of a neighbourhood of the core curve of one of the two solid tori that make up the lens space?
I figured out that I have to remove a solid torus from L(p,q) but I can not visulaize or proof what do I get if do that.
general-topology knot-theory
asked Dec 12 '18 at 13:50
ValVal
1
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1
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You get... the other solid torus.
$endgroup$
– user98602
Dec 12 '18 at 19:04
add a comment |
$begingroup$
I have given a lens space L(p,q). What compact three-manifold with boundary do we get if we remove the interior of a neighbourhood of the core curve of one of the two solid tori that make up the lens space?
I figured out that I have to remove a solid torus from L(p,q) but I can not visulaize or proof what do I get if do that.
general-topology knot-theory
asked Dec 12 '18 at 13:50
ValVal
1
$endgroup$
I have given a lens space L(p,q). What compact three-manifold with boundary do we get if we remove the interior of a neighbourhood of the core curve of one of the two solid tori that make up the lens space?
I figured out that I have to remove a solid torus from L(p,q) but I can not visulaize or proof what do I get if do that.
general-topology knot-theory
general-topology knot-theory
asked Dec 12 '18 at 13:50
ValVal
1
asked Dec 12 '18 at 13:50
ValVal
1
asked Dec 12 '18 at 13:50
ValVal
1
asked Dec 12 '18 at 13:50
ValVal
1
asked Dec 12 '18 at 13:50
ValVal
1
1
1
$begingroup$
You get... the other solid torus.
$endgroup$
– user98602
Dec 12 '18 at 19:04
add a comment |
1
$begingroup$
You get... the other solid torus.
$endgroup$
– user98602
Dec 12 '18 at 19:04
1
1
$begingroup$
You get... the other solid torus.
$endgroup$
– user98602
Dec 12 '18 at 19:04
$begingroup$
You get... the other solid torus.
$endgroup$
– user98602
Dec 12 '18 at 19:04
add a comment |
1 Answer
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It looks like the definition of $L(p,q)$ you are using is that you take two solid tori and identify their boundaries along a slope of $p/q$.
A solid torus minus its core is homeomorphic to $Ttimes I$, where $T=S^1times S^1$ and $I=[0,1]$. Every homeomorphism $Tto T$ extends to a homeomorphism $Ttimes Ito Ttimes I$ that is unique up to isotopy, with $T$ being identified with $Ttimes {0}$. Thus, there is only one way (up to homeomorphism) to identify the boundary of a solid torus with one of the boundary components of $Ttimes I$. The result is homeomorphic to a solid torus.
answered Dec 12 '18 at 19:34
Kyle MillerKyle Miller
9,058929
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$begingroup$
It looks like the definition of $L(p,q)$ you are using is that you take two solid tori and identify their boundaries along a slope of $p/q$.
A solid torus minus its core is homeomorphic to $Ttimes I$, where $T=S^1times S^1$ and $I=[0,1]$. Every homeomorphism $Tto T$ extends to a homeomorphism $Ttimes Ito Ttimes I$ that is unique up to isotopy, with $T$ being identified with $Ttimes {0}$. Thus, there is only one way (up to homeomorphism) to identify the boundary of a solid torus with one of the boundary components of $Ttimes I$. The result is homeomorphic to a solid torus.
answered Dec 12 '18 at 19:34
Kyle MillerKyle Miller
9,058929
$endgroup$
add a comment |
$begingroup$
It looks like the definition of $L(p,q)$ you are using is that you take two solid tori and identify their boundaries along a slope of $p/q$.
A solid torus minus its core is homeomorphic to $Ttimes I$, where $T=S^1times S^1$ and $I=[0,1]$. Every homeomorphism $Tto T$ extends to a homeomorphism $Ttimes Ito Ttimes I$ that is unique up to isotopy, with $T$ being identified with $Ttimes {0}$. Thus, there is only one way (up to homeomorphism) to identify the boundary of a solid torus with one of the boundary components of $Ttimes I$. The result is homeomorphic to a solid torus.
answered Dec 12 '18 at 19:34
Kyle MillerKyle Miller
9,058929
$endgroup$
add a comment |
$begingroup$
It looks like the definition of $L(p,q)$ you are using is that you take two solid tori and identify their boundaries along a slope of $p/q$.
A solid torus minus its core is homeomorphic to $Ttimes I$, where $T=S^1times S^1$ and $I=[0,1]$. Every homeomorphism $Tto T$ extends to a homeomorphism $Ttimes Ito Ttimes I$ that is unique up to isotopy, with $T$ being identified with $Ttimes {0}$. Thus, there is only one way (up to homeomorphism) to identify the boundary of a solid torus with one of the boundary components of $Ttimes I$. The result is homeomorphic to a solid torus.
answered Dec 12 '18 at 19:34
Kyle MillerKyle Miller
9,058929
$endgroup$
It looks like the definition of $L(p,q)$ you are using is that you take two solid tori and identify their boundaries along a slope of $p/q$.
A solid torus minus its core is homeomorphic to $Ttimes I$, where $T=S^1times S^1$ and $I=[0,1]$. Every homeomorphism $Tto T$ extends to a homeomorphism $Ttimes Ito Ttimes I$ that is unique up to isotopy, with $T$ being identified with $Ttimes {0}$. Thus, there is only one way (up to homeomorphism) to identify the boundary of a solid torus with one of the boundary components of $Ttimes I$. The result is homeomorphic to a solid torus.
answered Dec 12 '18 at 19:34
Kyle MillerKyle Miller
9,058929
answered Dec 12 '18 at 19:34
Kyle MillerKyle Miller
9,058929
answered Dec 12 '18 at 19:34
Kyle MillerKyle Miller
9,058929
answered Dec 12 '18 at 19:34
Kyle MillerKyle Miller
9,058929
9,058929
add a comment |
add a comment |
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You get... the other solid torus.
$endgroup$
– user98602
Dec 12 '18 at 19:04